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For a given reaction, presence of catalyst reduces the energy of activation by `2` kcal at 27^(@)C`. Thus rate of reaction will be increased by:A. `20` timesB. `14` timesC. `28` timesD. `2` times |
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Answer» Correct Answer - c `K_(1)/K_(2)=e^(-[E_(a1)-E_(a2)]//RT)=e^(-[(2xx10^(3))//(2xx300)])` `"log"_(e) K_(2)/K_(1)=(2xx10^(3))/(600)` `K_(2)/K_(1)=28=r_(2)/r_(1)` `:. R_(2)=28r_(1)` `=28` |
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